Self-testing of qutrit systems Jdrzej Kaniewski QMATH, Department - - PowerPoint PPT Presentation

Self-testing of qutrit systems

Jędrzej Kaniewski

QMATH, Department of Mathematical Sciences University of Copenhagen, Denmark

joint work with Antonio Acín, Remigiusz Augusiak, Flavio Baccari, Alexia Salavrakos, Ivan Šupić, Jordi Tura

CEQIP ’18 15 June 2018

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Bell nonlocality

Bell scenario j a k b P(a, b|j, k)

Bell nonlocality

Bell scenario j a k b P(a, b|j, k) Assume that P ∈ Q is quantum P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• .

Bell nonlocality

Bell scenario j a k b P(a, b|j, k) Assume that P ∈ Q is quantum P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• .

Def.: P ∈ L is local if P(a, b|j, k) =

• λ

p(λ) pA(a|j, λ) pB(b|k, λ). Bell: L Q ⇐ ⇒ “ quantum mechanics is (Bell) nonlocal ”

Bell nonlocality

Given some P ∈ Q, how to show that P ∈ L?

Bell nonlocality

Given some P ∈ Q, how to show that P ∈ L? Real vector C = (cabjk) define C, P :=

• abjk

cabjkP(a, b|j, k) and βL := max

P∈L C, P

(local value) βQ := max

P∈Q C, P

(quantum value) (suppose βL < βQ)

Bell nonlocality

Given some P ∈ Q, how to show that P ∈ L? Real vector C = (cabjk) define C, P :=

• abjk

cabjkP(a, b|j, k) and βL := max

P∈L C, P

(local value) βQ := max

P∈Q C, P

(quantum value) (suppose βL < βQ) Bell violation: C, P > βL = ⇒ P ∈ L

Bell nonlocality

Obs.: Separable states give local statistics (for all measurements) ρAB =

• λ

pλσλ ⊗ τλ,

Bell nonlocality

Obs.: Separable states give local statistics (for all measurements) ρAB =

• λ

pλσλ ⊗ τλ, P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• =
• λ

pλ · tr

• F j

aσλ

• pA(a|j,λ)

· tr

• Gk

bτλ

• pB(b|k,λ)

. Nonlocality = ⇒ entanglement

Bell nonlocality

Obs.: Separable states give local statistics (for all measurements) ρAB =

• λ

pλσλ ⊗ τλ, P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• =
• λ

pλ · tr

• F j

aσλ

• pA(a|j,λ)

· tr

• Gk

bτλ

• pB(b|k,λ)

. Nonlocality = ⇒ entanglement can we make this connection more explicit/rigorous?

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Self-testing

Given P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• deduce properties of ρAB, {F j

a}, {Gk b}

Self-testing

Given P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• deduce properties of ρAB, {F j

a}, {Gk b}

(i) we do not assume that ρAB is pure or that the measurements are projective (we want to rigorously deduce it!)

Self-testing

Given P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• deduce properties of ρAB, {F j

a}, {Gk b}

(i) we do not assume that ρAB is pure or that the measurements are projective (we want to rigorously deduce it!) (ii) often only promised some Bell violation C, P = β

Self-testing

Given P(a, b|j, k) = tr

• (F j

a ⊗ Gk b)ρAB

• deduce properties of ρAB, {F j

a}, {Gk b}

(i) we do not assume that ρAB is pure or that the measurements are projective (we want to rigorously deduce it!) (ii) often only promised some Bell violation C, P = β might seem like a hopeless task... ...but often can deduce essentially everything!

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Sum-of-squares decomposition

Given a Bell functional C, how to compute βQ = maxP∈Q C, P? Easy to provide lower bounds, what about upper bounds?

Sum-of-squares decomposition

Given a Bell functional C, how to compute βQ = maxP∈Q C, P? Easy to provide lower bounds, what about upper bounds?

1 Construct Bell operator

W :=

• abjk

cabjkF j

a ⊗ Gk b

Sum-of-squares decomposition

Given a Bell functional C, how to compute βQ = maxP∈Q C, P? Easy to provide lower bounds, what about upper bounds?

1 Construct Bell operator

W :=

• abjk

cabjkF j

a ⊗ Gk b

2 Prove that for all measurements

W ≤ c 1 for c ∈ R

Sum-of-squares decomposition

Given a Bell functional C, how to compute βQ = maxP∈Q C, P? Easy to provide lower bounds, what about upper bounds?

1 Construct Bell operator

W :=

• abjk

cabjkF j

a ⊗ Gk b

2 Prove that for all measurements

W ≤ c 1 for c ∈ R

3 Then βQ ≤ c because for all quantum realisations

C, P = tr(WρAB) ≤ c tr(ρAB) = c

Sum-of-squares decomposition

Q: How to show that W ≤ c 1 for all measurements?

Sum-of-squares decomposition

Q: How to show that W ≤ c 1 for all measurements? A: Write difference as sum of squares c 1 − W ≥

• j

L†

jLj.

(operators Lj depend on measurement operators) if βQ = c = ⇒ sum-of-squares (SOS) decomposition is tight

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Example: the CHSH inequality

the CHSH operator W := A0 ⊗ (B0 + B1) + A1 ⊗ (B0 − B1) where −1 ≤ Aj ≤ 1 and −1 ≤ Bk ≤ 1

Example: the CHSH inequality

the CHSH operator W := A0 ⊗ (B0 + B1) + A1 ⊗ (B0 − B1) where −1 ≤ Aj ≤ 1 and −1 ≤ Bk ≤ 1 define L0 = A0 ⊗ 1 − 1 ⊗ B0 + B1 √ 2 L1 = A1 ⊗ 1 − 1 ⊗ B0 − B1 √ 2

Example: the CHSH inequality

the CHSH operator W := A0 ⊗ (B0 + B1) + A1 ⊗ (B0 − B1) where −1 ≤ Aj ≤ 1 and −1 ≤ Bk ≤ 1 define L0 = A0 ⊗ 1 − 1 ⊗ B0 + B1 √ 2 L1 = A1 ⊗ 1 − 1 ⊗ B0 − B1 √ 2 check W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

Example: the CHSH inequality

the CHSH operator W := A0 ⊗ (B0 + B1) + A1 ⊗ (B0 − B1) where −1 ≤ Aj ≤ 1 and −1 ≤ Bk ≤ 1 define L0 = A0 ⊗ 1 − 1 ⊗ B0 + B1 √ 2 L1 = A1 ⊗ 1 − 1 ⊗ B0 − B1 √ 2 check W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

• W ≤ 2

√ 2 1 and βQ = 2 √ 2, so the SOS decomposition is tight

Example: the CHSH inequality

W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

• bserving tr(WρAB) = 2

√ 2 implies:

Example: the CHSH inequality

W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

• bserving tr(WρAB) = 2

√ 2 implies:

1 all measurements are projective on the local supports:

tr(A2

jρA) = tr(B2 kρB) = 1

Example: the CHSH inequality

W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

• bserving tr(WρAB) = 2

√ 2 implies:

1 all measurements are projective on the local supports:

tr(A2

jρA) = tr(B2 kρB) = 1

2 observables of Alice and Bob satisfy LjρAB = 0

(A0 ⊗ 1)ρAB =

• 1 ⊗ B0 + B1

√ 2

• ρAB

Example: the CHSH inequality

W = 1 √ 2

• (A2

0 + A2 1) ⊗ 1 + 1 ⊗ (B2 0 + B2 1) − (L† 0L0 + L† 1L1)

• bserving tr(WρAB) = 2

√ 2 implies:

1 all measurements are projective on the local supports:

tr(A2

jρA) = tr(B2 kρB) = 1

2 observables of Alice and Bob satisfy LjρAB = 0

(A0 ⊗ 1)ρAB =

• 1 ⊗ B0 + B1

√ 2

• ρAB

if ρA and ρB are full-rank, then A2

0 = 1 =

⇒ B0 + B1 √ 2 2 = 1 = ⇒ {B0, B1} = 0

Example: the CHSH inequality

the relation determines form of observables B2

0 = B2 1 = 1

and {B0, B1} = 0 = ⇒ B0 = UB(σx ⊗ 1)U †

B

B1 = UB(σz ⊗ 1)U †

B

Example: the CHSH inequality

the relation determines form of observables B2

0 = B2 1 = 1

and {B0, B1} = 0 = ⇒ B0 = UB(σx ⊗ 1)U †

B

B1 = UB(σz ⊗ 1)U †

B

the inequality is symmetric, so A0 and A1 have the same form construct W and determine the eigenspace corresponding to 2 √ 2

Example: the CHSH inequality

the relation determines form of observables B2

0 = B2 1 = 1

and {B0, B1} = 0 = ⇒ B0 = UB(σx ⊗ 1)U †

B

B1 = UB(σz ⊗ 1)U †

B

the inequality is symmetric, so A0 and A1 have the same form construct W and determine the eigenspace corresponding to 2 √ 2 Self-testing (rigidity) statement for CHSH: if β = 2 √ 2 then A0 = UA(σx ⊗ 1)U †

A

B0 = UB(σx ⊗ 1)U †

B

A1 = UA(σz ⊗ 1)U †

A

B1 = UB(σz ⊗ 1)U †

B

and ρAB = U(ΦA′B′ ⊗ τA′′B′′)U † for U := UA ⊗ UB

Example: the CHSH inequality

Strategy:

1 Find tight SOS decomposition 2 Deduce algebraic relations between local observables 3 Deduce their exact form (up to unitaries and extra degrees of

freedom)

4 Construct Bell operator and find eigenspace corresponding to βQ

Example: the CHSH inequality

Strategy:

1 Find tight SOS decomposition 2 Deduce algebraic relations between local observables 3 Deduce their exact form (up to unitaries and extra degrees of

freedom)

4 Construct Bell operator and find eigenspace corresponding to βQ

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Result 1: SATWAP inequality

maximally violated by maximally entangled state and CGLMP measurements (Remik’s talk) CGLMP measurement in dimension d for φ ∈ [0, 2π] |eφ

j :=

1 √ d

d−1

• k=0

ω(j−φ)k|k for ω := exp(2πi/d)

Result 1: SATWAP inequality

maximally violated by maximally entangled state and CGLMP measurements (Remik’s talk) CGLMP measurement in dimension d for φ ∈ [0, 2π] |eφ

j :=

1 √ d

d−1

• k=0

ω(j−φ)k|k for ω := exp(2πi/d) we look at 2 inputs and 3 outputs, optimal angles φ0 = 0, φ1 = 1/2; computing |e1/2

j′ |e0 j| gives

j\j′ 1 2 2/3 2/3 1/3 1 1/3 2/3 2/3 2 2/3 1/3 2/3 not mutually unbiased

Result 1: SATWAP inequality

SOS written in terms of observables (unitary for projective measurements) Aj =

• a

ωaF j

a

Bk =

• b

ωbGk

b

Result 1: SATWAP inequality

SOS written in terms of observables (unitary for projective measurements) Aj =

• a

ωaF j

a

Bk =

• b

ωbGk

b

SOS decomposition and some algebra = ⇒ projectivity and ω2B†

0 + ωB† 1 = −{B0, B1}

Result 1: SATWAP inequality

SOS written in terms of observables (unitary for projective measurements) Aj =

• a

ωaF j

a

Bk =

• b

ωbGk

b

SOS decomposition and some algebra = ⇒ projectivity and ω2B†

0 + ωB† 1 = −{B0, B1}

more algebra... = ⇒ B0, B1 are the CGLMP measurements acting on a qutrit (up to usual equivalences)

Result 1: SATWAP inequality

SOS written in terms of observables (unitary for projective measurements) Aj =

• a

ωaF j

a

Bk =

• b

ωbGk

b

SOS decomposition and some algebra = ⇒ projectivity and ω2B†

0 + ωB† 1 = −{B0, B1}

more algebra... = ⇒ B0, B1 are the CGLMP measurements acting on a qutrit (up to usual equivalences) construct Bell operator = ⇒ . . .

Result 1: SATWAP inequality

Result: Self-testing statement for SATWAP for d = 3

Result 1: SATWAP inequality

Result: Self-testing statement for SATWAP for d = 3

• Cor. 1: SATWAP functional has a unique maximiser

q u a n t u m s e t l

• c

a l s e t

• Cor. 2: The maximal violation certifies log 3 bits of local randomness

= ⇒ could use for cryptography

Result 1: SATWAP inequality

Result: Self-testing statement for SATWAP for d = 3

• Cor. 1: SATWAP functional has a unique maximiser

q u a n t u m s e t l

• c

a l s e t βL βQ P ∗

• Cor. 2: The maximal violation certifies log 3 bits of local randomness

= ⇒ could use for cryptography

Outline

Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Result 2: generalised CHSH inequality

j a k b P(a, b|j, k)

Result 2: generalised CHSH inequality

j a k b P(a, b|j, k) CHSH: a, b, j, k ∈ {0, 1} win ⇐ ⇒ a ⊕ b ⊕ jk = 0

Result 2: generalised CHSH inequality

j a k b P(a, b|j, k) CHSH: a, b, j, k ∈ {0, 1} win ⇐ ⇒ a ⊕ b ⊕ jk = 0 CHSHd: a, b, j, k ∈ {0, 1, . . . , d − 1} win ⇐ ⇒ a + b + jk ≡ 0 mod d

Result 2: generalised CHSH inequality

Buhrman and Massar (’05) proposed and studied d = 3 Ji et al. (’08) and Liang et al. (’09) studied higher d (mainly prime)

Result 2: generalised CHSH inequality

Buhrman and Massar (’05) proposed and studied d = 3 Ji et al. (’08) and Liang et al. (’09) studied higher d (mainly prime) inconclusive! classical value, quantum value, optimal realisation: not understood

Result 2: generalised CHSH inequality

Buhrman and Massar (’05) proposed and studied d = 3 Ji et al. (’08) and Liang et al. (’09) studied higher d (mainly prime) inconclusive! classical value, quantum value, optimal realisation: not understood conclusion: this Bell functional seems natural, but turns out to be ill-behaved

Result 2: generalised CHSH inequality

Buhrman and Massar (’05) proposed and studied d = 3 Ji et al. (’08) and Liang et al. (’09) studied higher d (mainly prime) inconclusive! classical value, quantum value, optimal realisation: not understood conclusion: this Bell functional seems natural, but turns out to be ill-behaved

Result 2: generalised CHSH inequality

Buhrman and Massar (’05) proposed and studied d = 3 Ji et al. (’08) and Liang et al. (’09) studied higher d (mainly prime) inconclusive! classical value, quantum value, optimal realisation: not understood conclusion: this Bell functional seems natural, but turns out to be ill-behaved

Result 2: generalised CHSH inequality

Bell operator reads Wd := 1 d3

d−1

• n=0

d−1

• j,k=0

ωnjkAn

j ⊗ Bn k

Result 2: generalised CHSH inequality

Bell operator reads Wd := 1 d3

d−1

• n=0

d−1

• j,k=0

ωnjkAn

j ⊗ Bn k

we consider prime d and W ′

d := 1

d3

d−1

• n=0

λn,d

d−1

• j,k=0

ωnjkAn

j ⊗ Bn k

for λn,d ∈ C, |λn,d| = 1

Result 2: generalised CHSH inequality

Bell operator reads Wd := 1 d3

d−1

• n=0

d−1

• j,k=0

ωnjkAn

j ⊗ Bn k

we consider prime d and W ′

d := 1

d3

d−1

• n=0

λn,d

d−1

• j,k=0

ωnjkAn

j ⊗ Bn k

for λn,d ∈ C, |λn,d| = 1 for the right choice of λn,d the quantum value can be computed analytically (tight SOS decomposition)!

Result 2: generalised CHSH inequality

quantum realisation achieving the quantum value |Φ = 1 √ d

d−1

• j=0

|j|j Bk = ωk(k+1)XZk Aj = . . . the observables correspond to d distinct bases which are pairwise mutually unbiased for d = 3 SOS relations allow us to prove self-testing!

Result 2: generalised CHSH inequality

For d = 3 the phases are: λ0,d = 1, λ1,d = e−iπ/18, λ2,d = e+iπ/18 (e−iπ/18 ≈ 0.9849 − 0.1737i ≈ 1) SOS + algebra = ⇒ projectivity and B†

0 = −ω{B1, B2}

(and perm.) this is sufficient to deduce the form of observables. . .

Result 2: generalised CHSH inequality

For d = 3 the phases are: λ0,d = 1, λ1,d = e−iπ/18, λ2,d = e+iπ/18 (e−iπ/18 ≈ 0.9849 − 0.1737i ≈ 1) SOS + algebra = ⇒ projectivity and B†

0 = −ω{B1, B2}

(and perm.) this is sufficient to deduce the form of observables. . . . . . except that now there are two inequivalent solutions (B0, B1, B2) ≡ (BT

0 , BT 1 , BT 2 )

not unitarily equivalent (σx, σy, σz) ≡ (σx, −σy, σz) local Hilbert spaces decompose into the “right-” and “left-handed” subspace and maximal violation possible only if Alice and Bob use opposite types!

Result 2: generalised CHSH inequality

Maximal violation certifies: |Φ =

1 √ 3

2

j=0 |j|j

specific MUB measurements for each party the two measurements must be of the opposite type

Result 2: generalised CHSH inequality

Maximal violation certifies: |Φ =

1 √ 3

2

j=0 |j|j

specific MUB measurements for each party the two measurements must be of the opposite type Corollaries: has unique maximiser in Q certifies log 3 bits of local randomness

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test proposed a family of Bell inequalities maximally violated by the maximally entangled state and MUB measurements

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test proposed a family of Bell inequalities maximally violated by the maximally entangled state and MUB measurements for d = 3 this a self-test (right/left-handed twist!)

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test proposed a family of Bell inequalities maximally violated by the maximally entangled state and MUB measurements for d = 3 this a self-test (right/left-handed twist!) self-testing results not relying on self-testing of qubit subspaces

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test proposed a family of Bell inequalities maximally violated by the maximally entangled state and MUB measurements for d = 3 this a self-test (right/left-handed twist!) self-testing results not relying on self-testing of qubit subspaces Open questions: extend SATWAP self-testing to arbitrary d extend generalised CHSH self-testing to arbitrary prime d

Conclusions and open questions

Conclusions: SATWAP inequality for d = 3 is a self-test proposed a family of Bell inequalities maximally violated by the maximally entangled state and MUB measurements for d = 3 this a self-test (right/left-handed twist!) self-testing results not relying on self-testing of qubit subspaces Open questions: extend SATWAP self-testing to arbitrary d extend generalised CHSH self-testing to arbitrary prime d robustness!

So you can really certify quantum systems without trusting the devices at all? Yes, Pooh, quantum mechanics is very strange and nobody really understands it, but let’s talk about it another day. . .